Question

An element crystallizes in a structure having FCC unit cell of an edge 200 pm. Calculate the density if 200 gm of it contains $24\times {10}^{23}$ atoms:

• A. $41.6gm/c{m}^3$
• B. $42‘.6gm/c{m}^3$
• C. $43.6gm/c{m}^3$
• D. $44.6gm/c{m}^3$

Answer 1

The correct option is A $41.6gm/c{m}^3$

Given,

Edge length$=200pm$

Number of atoms$=24\times {10}^{23}$

Given weight of element$=200g$

Structure of the element = Face centered cubic

In fcc unit cell there are $4$ atoms per unit cell.

We know that $Density=\frac{Mass}{Volume}$ $...\left(i\right)$

Mass of the unit cell$=\frac{200\times 4}{24\times {10}^{23}}$

$\Rightarrow 33.3\times {10}^{-23}g$

Volume of the cubic unit cell$=(edge{)}^3$

$\Rightarrow (200\times {10}^{-10}{)}^3$

$\Rightarrow 8\times {10}^{-24}c{m}^3$

Substituting the value of mass and volume in equation$\left(i\right)$, we get

$Density=\frac{33.3\times {10}^{-23}g}{8\times {10}^{-24}c{m}^3}$

$\Rightarrow 41.6gc{m}^{-3}$

Therefore, the density of the element will be $41.6gc{m}^{-3}$